Solve it using the given values provided below, The efficiency must be 35% In a simple rankine cycle, the steam throttled condition is 8 MPa and 480°C. The steam is then reheated to 2 MPa and 460 °C. If turbine exhaust is 60 °C, determine the following: a. Heat Added b. Turbine Work c. Heat Rejected d. Pump Work e. Net Work f. Efficiency
Solve it using the given values provided below, The efficiency must be 35% In a simple rankine cycle, the steam throttled condition is 8 MPa and 480°C. The steam is then reheated to 2 MPa and 460 °C. If turbine exhaust is 60 °C, determine the following: a. Heat Added b. Turbine Work c. Heat Rejected d. Pump Work e. Net Work f. Efficiency
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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Solve it using the given values provided below,
The efficiency must be 35%
In a simple rankine cycle, the steam throttled condition is 8 MPa and 480°C. The steam is then reheated to 2 MPa and 460 °C. If turbine exhaust is 60 °C, determine the following:
a. Heat Added
b. Turbine Work
c. Heat Rejected
d. Pump Work
e. Net Work
f. Efficiency
![8 MPa and 480 °C (Table 3. Vapor) page 72
h: = 3348.4 kJ/kg
S1 = 6.6586 kJ/kg – K
O 2 MPa (Table 3. Vapor; note: s1 = s2) page 46
h2 = 2963.145 kJ/kg
2 MPa and 460 °C (Table 3. Vapor; note: s3 = Sa) page 46
ha = 3379.5 k/kg
S3 = 7.3147 kJ/kg –K](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe0e13e51-6dee-4752-a6c2-df0b0f97152c%2F7de29f90-87fc-438d-99dd-79bbc75a2e8b%2Fg18krwx_processed.jpeg&w=3840&q=75)
Transcribed Image Text:8 MPa and 480 °C (Table 3. Vapor) page 72
h: = 3348.4 kJ/kg
S1 = 6.6586 kJ/kg – K
O 2 MPa (Table 3. Vapor; note: s1 = s2) page 46
h2 = 2963.145 kJ/kg
2 MPa and 460 °C (Table 3. Vapor; note: s3 = Sa) page 46
ha = 3379.5 k/kg
S3 = 7.3147 kJ/kg –K
![60°C (Table 1. Saturation Temperatures) page 3
Ps
= 0.019940 MPa
19.94 kPa
hy
= 251.13 kJ/kg.
hs
= 2358.5 kJ/kg
= 0.8312 kJ/kg – K
= 7.0784 kl/kg – K
V
= 0.0010172 m/kg
Vs](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe0e13e51-6dee-4752-a6c2-df0b0f97152c%2F7de29f90-87fc-438d-99dd-79bbc75a2e8b%2F6nokuz_processed.jpeg&w=3840&q=75)
Transcribed Image Text:60°C (Table 1. Saturation Temperatures) page 3
Ps
= 0.019940 MPa
19.94 kPa
hy
= 251.13 kJ/kg.
hs
= 2358.5 kJ/kg
= 0.8312 kJ/kg – K
= 7.0784 kl/kg – K
V
= 0.0010172 m/kg
Vs
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